Question by Alex: How many grams of hydrogen can be stored in each tank?
An experimental car uses hydrogen as a fuel and has a tank which can hold gaseous hydrogen. The designers have two options: one tank is 42 L in size and maintained at ambient temperature (i.e. 25 C), the other tank is 28 L in size, but maintained at 0oC. Both tanks are pressurized to 5.00 atm. How many grams of hydrogen can be stored in each tank?
how many grams of water will be produced if the car burns 15.5 L of hydrogen from teach tank?

Best answer:

Answer by sfgrc
To determine the grams of hydrogen stored in each tank you can use the Ideal Gas Law:

PV=nRT where

P=pressure of the gas
V=volume of the gas
n=number of moles of the gas
R=universal gas constant (0.08205lit-atm/mol-K)
T=absolute temperature of the gas

For the 42L tank:

PV=nRT

=> n=PV/RT

=(5.00atm)*(42L)/((0.08205lit-atm/mol-K)*(25+273K))

=8.58 moles H2 gas

You get grams by multiplying by the molecular weight (2.01g/mole H2):

g H2 = (8.58moles H2)*(2.01g/mole H2) = 17.25g H2 in the 42L tank.

You can do the same thing for the 28L tank:

n=PV/RT = (5.00atm)*(28L)/((0.08205lit-atm/mol-K)*(0+273K))

=6.25 moles H2

g H2 = (6.25 moles)*(2.01g/mol) = 12.6 g H2 in the 28L tank.

For the last part of the question, I will assume the 15.5 L of hydrogen is measured at standard conditions T=0C P=1atm

n=PV/RT

=(1atm)*(15.5L)/((0.08205lit-atm/mol-K)*(0+273K))

=0.692 moles H2

If you write the balanced chemical reaction for the burning of H2 you have:

H2 + 1/2 O2 -> H2O

and you can see that 1 mole of H2O is formed for every 1 mole of H2 consumed, so

moles H2O = moles H2 = 0.692 moles H2O

You get the grams of H2O by multiplying by its molecular weight (18.0g/mol):

g H2O = (0.692moles)*(18.0g/mole) = 12.5 gH2O

If the 15.5 L is intended to be measured at the pressure of each tank, then you would follow the same procedure, except use the appropriate pressures.

Hope this helps.

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